Answer to Question #223852 in Statistics and Probability for Pelumi

Question

A card is drawn at random from an ordinary deck of 52 playing cards. Describe the sample space if consideration

of suits (a) is not, (b) is, taken into account.

Referring to the experiment of Problem 1 above, let A be the event {king is drawn} or simply {king} and B the event {club is drawn} or simply {club}. Describe the events (a) AᵁB, (b) A ՈB, (c) AᵁB’, (d) A’ᵁB’,

1. (e) A - B, (f) A’- B’, (g) (AՈB) υ (AՈB’).

Answer 1

Solution:

(a) If we do not take into account the suits, the sample space consists of ace, two, ..., ten, jack, queen, king, and it can be indicated as $\{1,2, \ldots, 13\}$ .

(b) If we do take into account the suits, the sample space consists of ace of hearts, spades, diamonds, and clubs; $\ldots$ ; king of hearts, spades, diamonds, and clubs. Denoting hearts, spades, diamonds, and clubs, respectively, by 1,2,3,4, for example, we can indicate a jack of spades by (11,2). The sample space then consists of the 52 points shown in Fig. 1.

(a) $A \cup B=\{ either\ king\ or\ club (or\ both, i.e., king\ of\ clubs)\}$ .

(b) $A \cap B=\{ both\ king\ and\ club \}=\{ king\ of\ clubs \}$ .

(c) Since $B=\{ club \}, B^{\prime}=\{ not\ club \}=\{ heart, diamond, spade \}$ .

Then A $\cup B^{\prime}=\{ king\ or\ heart\ or\ diamond\ or\ spade \}$ .

(d) $A^{\prime} \cup B^{\prime}=\{ not\ king\ or\ not\ club \}=\{ not\ king\ of\ clubs \}=\{ any\ card\ but\ king\ of\ clubs \}$ . This can also be seen by noting that $A^{\prime} \cup B^{\prime}=(A \cap B)^{\prime}$ and using (b).

(e) $A-B=\{ king\ but\ not\ club\}$ .

This is the same as $A \cap B^{\prime}=\{ king\ and\ not\ club \}$ .

(f) "A^{\\prime}-B^{\\prime}=\\{ not\\ king\\ and\\ not\\ "not\\ club" \\}=\\{ not\\ king\\ and\\ club \\}=\\{ any\\ club\\ except\\ king \\}" . This can also be seen by noting that $A^{\prime}-B^{\prime}=A^{\prime} \cap\left(B^{\prime}\right)^{\prime}=A^{\prime} \cap B$ .

(g) $(A \cap B) \cup\left(A \cap B^{\prime}\right)=\{ (king\ and\ club) or (king\ and\ not\ club) \}=\{ king \}$ . This can also be seen by noting that $(A \cap B) \cup\left(A \cap B^{\prime}\right)=A$ .